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titofigueroa777
08.06.2020 •
Mathematics
How much is one gram into ounces?
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Ответ:
It would be about 0.035274
Step-by-step explanation:
For an approximate result, divide the mass value by 28.35
Hope this helps
Ответ:
the time of death td= 9:42 am
Step-by-step explanation:
If we suppose that a mass of constant heat capacity is cooled at a rate that is proportional to the temperature difference between the body and the environment then we can use an exponencial model for the body temperature as
T - T env = a*e^(-t/b)
then
if we use the reference t=0 as 10:30 am , then at t=0 , T= 76.3°F and
T - T env = a*e^(-t/b)
76.3°F-64°F = a*e^(-0/b)
a = 76.3°F-64°F = 12.3 °F
then for t= 0.5 hours (11:00 am from 10:30 am) ,T= 72.7°F and
T - T env = a*e^(-t/b)
72.7°F - 64°F = 12.3 °F*e^(-0.5 h /b)
b = -0.5 h/ln [( 72.7°F - 64°F) / 12.3 °F ] = 1.44 h
finally for Td=85.3°F the time of death td= is
Td - T env = a*e^(-td/b)
td= -b* ln[(Td - T env)/a]
td= - 1.44 h * ln[(85.3 °F - 64°F )/12.3 °F ] ]= - 0.79 h from 10:30 am
therefore
td= - 0.79 h*60 min/hour= -47.4 min from 10:30 am = 9:42 am
Note :
we applied the following physical model to end with the exponencial function
Q= m*c * (T-To) → dQ/dt= m*c*dT/dt
and -dQ/dT= U*A*(T- T env)
then
m*c*dT/dt = -U*A*(T- T env)
-dT/dt = U*A/(m*C) *(T - T env)
ln ( T- T env) = U*A/(m*C) (-t) + C
T- T env = e^(U*A/(m*C) t + C) = e^[U*A/(m*C)*(-t)] *e^C
T- T env = a*e^(-t/b)